If the lines \frac{x-1}{-3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}\: \: and\: \: \frac{x-1}{3\lambda}=\frac{y-1}{2}=\frac{z-6}{-5}  are perpendicular, find the value of \lambda. Hence find whether the lines are intersecting or not.

 

 

 

 

 
 
 
 
 

Answers (1)

The equation of the given lines are:

\frac{x-1}{-3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}\: \: and\: \: \frac{x-1}{3\lambda}=\frac{y-1}{2}=\frac{z-6}{-5}

If these lines are perpendicular, then

a_1a_2+b_1b_2+c_1c_2=0

\Rightarrow -3 \times 3\lambda +2\lambda \times 2+2\times \left ( -5 \right )=0

\Rightarrow -9\lambda +4\lambda -10=0

\Rightarrow -5\lambda =10

\lambda =-2

Now the lines are

\frac{x-1}{-3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}\: \: and\: \: \frac{x-1}{3\lambda}=\frac{y-1}{2}=\frac{z-6}{-5}The coordinates of any point on first-line are given by:

\frac{x-1}{-3}=\frac{y-2}{-4}=\frac{z-3}{2}=\alpha

or x-1=-3\alpha \Rightarrow x=-3\alpha +1

y-2=-4\alpha \Rightarrow y=-4\alpha +2

z-3=2\alpha \Rightarrow z=2\alpha +3

So, coordinates of any point on this line are (-3\alpha+1, 4\alpha+2, 2\alpha+3)

The coordinates of any point on the second line are given by:

\frac{x-1}{-6}=\frac{y-1}{2}=\frac{z-6}{-5}=\beta

or x-1=-6\beta \Rightarrow x=-6\beta +1

y-1=2\beta \Rightarrow y=2\beta+1 z-6=-5\beta \Rightarrow z=-5\beta+6

So, coordinates of any point on second line are \left ( -6\beta+1, 2\beta+1, -5\beta+6 \right )

If lines intersect then they have a common point. So far some values of \alpha \: \: and\: \: \beta we have

-3\alpha+1= -6 \beta +1\Rightarrow -3\alpha= -6 \beta

\Rightarrow \alpha= 2 \beta

and \Rightarrow -4\alpha+2= 2 \beta +1

\Rightarrow -4\alpha+1= 2 \beta

On solving, we have 

\alpha =\frac{1}{5}\: \: \: and\: \: \beta=\frac{1}{10 }

The values of \alpha \: and\: \beta do not satisfy the third equation. 

Hence, lines do not intersect with each other.

 

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